1. Information and Computer Security CPIS 312 Lab 4 & 5
Symmetric key
cryptography
TRIGUI Mohamed Salim

2. Lab Objectives
Students differentiate between stream cipher and block cipher.
Students understand what is LSFR
Students will learn how A5/1 algorithm works.
Students will try to apply A5/1 algorithm practically.
Students will learn how RC4 algorithm works.
Students will try to apply RC4 algorithm practically.

3. Lab Outcomes
At the end of this lab, students will be able to work with A5/1 algorithms as example of stream cipher by testing it practically during the lab.
At the end of this lab, students will be able to work with RC4 algorithms as example of stream cipher by testing it practically during the lab.

4. Stream cipher
A stream cipher is a symmetric cipher where convert one symbol of Plaintext immediately into a symbol of Ciphertext.
Algorithms:
Use one of the algorithms to generate the Key Stream (S) from the Key (K)
C = P  S; where C is the ciphertext, and P is the plaintext

5. Block cipher Encrypts a group of plaintext symbols as one block.
It works on blocks of plaintext and produce blocks of ciphertext
The columnar transposition is an example of block ciphers
A block cipher might take a 128-bit block of plaintext as input, and output.

6. Feedback Function : XOR
LFSR structure
A5/1 based on Linear Feedback Shift Registers Purpose - to produce pseudo random bit sequence Consists of two parts : shift register – bit sequence feedback function Tap Sequence : bits that are input to the feedback function b1 b2 b3 b4
...
bn-1 bn
output
new value
Feedback Function : XOR

7. LFSR features
LFSR Period – the length of the output sequence before it starts repeating itself.
n-bit LFSR can be in 2n-1 internal states
 the maximal period is also 2n-1
the tap sequence determines the period
the polynomial formed by a tap sequence plus 1 must be a primitive polynomial (mod 2)
What is primitive polynomial?
Boolean polynomial p(x) that can be used to compute the increasing powers of n of x^n mod p(x), to obtain all possible non-zero polynoomials of degree less than p(x)

8. LFSR example x12+x6+x4+x+1 corresponds to LFSR of length 12 b1 b2 b3

9. A5/1 A5/1 consists of 3 shift registers X: 19 bits (x18,x17,x16, …,x0)
Y: 22 bits (y21,y20,y19, …,y0)
Z: 23 bits (z22,z21,z20, …,z0)

10. A5/1 At each step: m = maj(x8, y10, z10) If x8 = m then X steps
Examples: maj(0,1,0) = 0 and maj(1,1,0) = 1
If x8 = m then X steps
t = x18  x17  x16  x13
xi = xi1 for i = 18,17,…,1 and x0 = t
If y10 = m then Y steps
t = y21  y20
yi = yi1 for i = 21,20,…,1 and y0 = t
If z10 = m then Z steps
t = z22  z21  z20  z7
zi = zi1 for i = 22,21,…,1 and z0 = t
Keystream bit is x18  y21  z22

11. A5/1 Each value is a single bit Key is initial fill of registerX
x18
x17
x16
x15
x14
x13
x12
x11
x10 x9 x8 x7 x6 x5 x4 x3 x2 x1 x0  Y 
y21
y20
y19
y18
y17
y16
y15
y14
y13
y12
y11
y10 y9 y8 y7 y6 y5 y4 y3 y2 y1 y0  Z
z22
z21
z20
z19
z18
z17
z16
z15
z14
z13
z12
z11
z10 z9 z8 z7 z6 z5 z4 z3 z2 z1 z0 
Each value is a single bit
Key is initial fill of register
Each register steps or not, based on (x8, y10, z10)
Keystream bit is XOR of left bit of each register

12. A5/1 Example Each register steps or not, based on (x8, y10, z10)1  Y  1  1 Z 
Each register steps or not, based on (x8, y10, z10)
Keystream bit is XOR of right bits of registers
Each register element is a single bit
Key is initial fill of register

13. A5/1 Example We have m = maj(0,1,1) = 1    Y  Z 1   Y 1  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1  0 = 1   1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Z
1  1  0  1 = 1 
We have m = maj(0,1,1) = 1
Register X does nothing
Registers Y and Z step
Keystream bit is 0  0  1 = 1

14. RS4 Invented by Ron Rivest
“RC” is “Ron’s Code” or “Rivest Cipher”
A stream cipher Generate keystream byte at a step:
Efficient in software
Simple and elegant
Used lots of places: WEP, etc.
Most popular stream cipher in existence

15. RS4 algorithm Two phases Initialization (Key Setup)
f = ( f + Si+ Kg) mod 256
Swap Si with Sf
key stream byte (Ciphering (XOR))
i=f=0
i = ( i + 1) mod 256 & f = ( f + Si) mod 256
t= (Si + Sf ) mod 256
KeystreamByte = S[t]

16. We use 4 bytes state, and 2 bytes key
RS4 example
We use 4 bytes state, and 2 bytes key
Initialization (Key Setup)
Iteration 1
i=f=g=0
S[]=[s0, s1, s2, s3]= [0, 1, 2, 3]
K[]= [k0, k1]= [2, 5]
f = (f + S0+ K0) mod 4
f = ( ) mod 4=2
Then Swap S0 with S2
New array S[]=[s0, s1, s2, s3]= [2, 1, 0, 3]
i=0+1=1
g=(0+1) mod 2=1

17. RS4 example
Iteration 2 i=1, f=2, g=1 S[]=[s0, s1, s2, s3]= [2, 1, 0, 3] K[]= [k0, k1]= [2, 5] f = (f + S1+ K1) mod 4 f = ( ) mod 4=0 Then Swap S1 with S0 New array S[]=[s0, s1, s2, s3]= [1, 2, 0, 3] i=1+1=2 g=(1+1) mod 2=0

18. RS4 example
Iteration 3 i=2, f=0, g=0 S[]=[s0, s1, s2, s3]= [1, 2, 0, 3] K[]= [k0, k1]= [2, 5] f = (f + S2+ K0) mod 4 f = ( ) mod 4=2 Then Swap S2 with S2 New array S[]=[s0, s1, s2, s3]= [1, 2, 0, 3] i=2+1=3 g=(0+1) mod 2=1

19. RS4 example
Iteration 4 i=3, f=2, g=1 S[]=[s0, s1, s2, s3]= [1, 2, 0, 3] K[]= [k0, k1]= [2, 5] f = (f + S3+ K1) mod 4 f = ( ) mod 4=2 Then Swap S3 with S2 New array S[]=[s0, s1, s2, s3]= [1, 2, 3, 0]

20. RS4 example Our plaintext is: HI key stream byte “H” i=f=0
S[]=[s0, s1, s2, s3]= [1, 2, 3, 0]
i = (i + 1) mod 4
i = (0 + 1) mod 4=1
f = (f + si) mod 4
f = (0 + 2) mod 4=2
Then Swap S1 with S2
New array S[]=[s0, s1, s2, s3]= [1, 3, 2, 0]

21. RS4 example
t= (S1 + S2 ) mod 4 t= (3 + 2 ) mod 4=1 Key stream Byte = S[1]=3 ( ) H XOR

22. RS4 example Our plaintext is: HI “I” i=1, f=2
S[]=[s0, s1, s2, s3]= [1, 3, 2, 0]
i = (i + 1) mod 4
i = (1 + 1) mod 4=2
f = (f + si) mod 4
f = (2 + 2) mod 4=0
Then Swap S2 with S0
New array S[]=[s0, s1, s2, s3]= [2, 3, 1, 0]

23. RS4 example
t= (S2 + S0 ) mod 4 t= (1 + 2 ) mod 4=3 Key stream Byte = S[3]=0 ( ) H XOR Plaintext: Ciphertext: